Paper I Material: Facts of Indian Constitution

Interesting Facts of Indian Constitution

ü    The Constitution of India is written and is the most elaborate constitution of the world. It was finally passed and accepted on Nov 26, 1949.

ü    The National Emblem was adopted by the Government on 26 January, 1950.

ü    There are 12 schedules, 448 articles, and 98 amendments, till date, in our Constitution.

ü    The words ‘SOCIALIST’, ‘SECULAR’ and ‘UNITY’ & ‘INTEGRITY’ were added to its preamble by the 42nd Amendment in 1976.

ü    Preamble is not justifiable.

ü    Indian Constitution is a rigid one. An amendment has to be passed by at least two-thirds majority of the members present and voting there in. However, in addition to this process, some amendments must be approved by at least 50% of the states. After this procedure the amendment is signed by the head of the state i.e; the President of India.

ü    There is a clear division of powers. Our constitution enumerates three lists, viz. the Union, the State and the Concurrent List. 

ü    The Union List consists of 97 subjects of national importance such as Defence, Railways, Post and Telegraph, etc. The State List consists of 66 subjects of local interest such as Public Health, Police etc. The Concurrent List has 47 subjects important to both the Union and the State. such as Electricity, Trade Union, Economic and Social Planning, Education etc.

ü    Both the Parliament and the State Legislatures can make laws on the subjects included in Concurrent list.

ü    The State governments may delegate some of its administrative functions relating to the State subjects, to Union Government for a specified period.

ü    The Indian states have no freedom to secede or separate from the Union. The federation is a union and Constitution is not truly a federal one. For example, The federal principle envisages a dual system of Courts. But, in India we have unified Judiciary with the Supreme Court at the apex.

ü    Constitutional experts have called it ‘semi-federal’ of ‘quasi federal’ system.

ü    The Centre appoints the Governors of the States and may take over the administration of the State on the recommendations of the Governor or otherwise.

ü    State Governors act more as Centre’s representative than as the head of the State. This enables the Union government to exercise control over the State administration.

ü    Indian states have equal representation in Lok Sabha but unequal representation in the Rajya Sabha.

ü    The States cannot propose amendments to, the Constitution. As such amendments can only be made by the Union Parliament.

ü    All important appointments such as the Chief Election Commissioner, the Comptroller and Auditor General are made by the Union Government.

ü    The Parliament, by law may increase or decrease the area of any State and may alter its name and boundaries.

ü    The Sarkaria Commission was appointed by Union Government to suggest ways and means to improve

ü    Centre-State relations. The commission recommended the establishment of permanent Inter-State Council.

7 Steps to success in UGC NET Exam

Hello Friends!

Cracking UGC NET examination, especially in a subject like Computer Science, is not a piece of cake. But at the same time, it is not an uphill task, which only selected few in the world can do.

Many of the candidates, after failing in 1-2 attempts, start thinking that it is not their cup of tea. I too failed to crack the exam, not just once or twice, but even after 7 attempts. However, my misery lasted only a few days. Every time the notification for UGC NET was out, I always decided to give it a shot. I must admit that every time I missed the sweet success, I felt low on confidence. Still I had a gut feeling that I was doing better than my last attempts and inch by inch, I was getting closer towards my goal.

Now, after qualifying UGC NET exam twice, I have gained more confidence in my teaching abilities. I have also realised that the way I planned and studied for getting success in NET exam was a kind of mixed bag. Means, if I had planned in a better manner and had devoted more time to practice, then it cannot take one more than 2 years to crack the exam!!

Here, I want to share with you my experience and my topsy-turvy journey towards this goal. If you do not want to spend 3-4 years to qualify this exam (like I did) and qualify the exam in less than two years, then, get ready to follow the steps (with determination) I am mentioning here:

Step 1:
Buy/procure the standard text books for different subjects plus one or two good MCQ books on computer Sci. (At the end, I will list names of some of good books). You need not to buy books on all subjects, focus on maximum 9-10 subjects, since you are not supposed to answer every question right and you are never expected to score 90% and above ;-)

Step 2:
Download all previous question papers (of all 3 papers) from the following website links of UGC:

http://ugcnetonline.in/news.php
http://www.cbsenet.nic.in
http://www.ugc.ac.in/net/oldqpyr.aspx?sub=87
http://www.ugc.ac.in/net/oldqpyr.aspx?sub=00

Also download all available answer keys of different exams (from June' 12 to June' 14, you will find ans keys of all 3 papers at UGCNetonline site, for NET exam December 2014 onward, please check CBSENET site).

DO TAKE PRINTOUT of all papers. It is better to use printed paper than reading it on your screen. However, to save your time and energy, better focus on question papers of last 4-5 years only and leave very old papers. For example, patter of question papers of 2004 and 2005 is very old and not many of the questions of the old pattern are asked nowadays.

Step 3:
Study the questions of previous papers thoroughly. Remember, this step is very important to understand where you stand now and how much time and preparation you will need. So, do this religiously and devote next 2-3 days in reading the questions and try to understand what is asked by the examiner. You will surely gain a good understanding about 3 things:

1. What are the subjects I must focus on
2. What are the subjects I can ignore or I can't study
3. What are the important topics of each subject from the point of view of NET exam

There are definitely some subjects of you will have a weak understanding. Rather than wasting your time and energy on these topics, better you focus on strengthening the subjects you have a flair as well as you can prepare. For example, I was not comfortable with Image Processing, Computer Graphics and ANN (Artificial Neural Network), so I left them out. For such topics, one would be tempted to not to miss them. I have a tip for these. For questions beyond your comprehension, just memorise their right answer. Who knows, the same question is asked in next exam :-)

Step 4:
Select one or two subjects (at max) to begin with and start reading their text book. For next 7-10 days, read only the books on the selected 1-2 topics. ALWAYS KEEP A HIGHLIGHTER OR A GOOD PENCIL with you. Any statement or concept which you may find important, then immediately highlight the points in the book itself. (You will be able to do this easily if you did step 3 right)

After finishing the book, go to
step 5. After step 5 is finished, come back to this step and follow the similar exercise for next 1-2 topics in the sequence.

Step 5:
Now, start solving the question papers you downloaded earlier. (Here, you need to solve only questions of the subject you studied in step 4). Eureka!! You will find that you are able to understand the question better now than previously-before reading the text book. (Still, there will be many questions you will not be able to solve). Mark your answers with a pencil then compare your answers with official ans. keys of UGC. For all the questions, you were not able to solve, go to step 6:

Step 6:
Research the tricky questions on Net (Internet dear, not UGC NET ;-) ). Take the help of Google Baba as well as your Facebook group of UGC to find the solution of the question. Stay determined to find the solution till you have not found the satisfactory explanation of the question). Who knows if the same question is asked in the exam itself, then you will be happy for your homework.

Step 7:
The final step is highly important-perseverance. Come what may, stay determined to get through your goal at any cost. Follow all 6 steps, especially step 5 and 6 in a loop, with all your determination. If you fail in one attempt, hold your emotions and bounce back to fight for the next exam.

After taking exam of one session, say June, do not sit idle and wait for the result. Continue with other subjects, solve the questions of your last exam. If you were not able to succeed, you will be left with lesser time to prepare for the exam of the next session. If you are already following your routine, nothing will be unexpected for you.

Stay focused, determined and relaxed also. Enjoy the feeling of putting your best foot forward. Pray to God to keep you on track. Nobody can stop you from achieving your share of success and joys in this world.

ALL THE BEST!!

Useful Q & A on Data Communication & Networks - Part 2

Hello Friends,

Based on the types of questions being asked in UGC NET, I have compiled some important questions on the topic of Data Communication & Networks. Please not just read this post, but also understand the concept behind the given answers. I have given explanation of these questions to help this process simple for you. Since Data Communication & Networks is one of the subjects on which lots of questions are asked, do make a note of preparing this subject well.

Please share your ideas and doubts also after reading the post.......

Q 1. What is the bucket size, when the overlapping and collision occur at same time?

Ans.: If there is only one entry possible in the bucket, when the collision occurs, there is no way to accommodate the colliding value. This results in the overlapping of values.

Q 2. Using RSA algorithm, what is the value of cipher text C, if the plain text M = 5 and p=3, q=11 & d=7 ?

A. 33            B. 5              C. 25            D. 26

Here p and q will be used to calculate public keys. Steps for transmission:

1. calculate n = p * q = 3 * 11 = 33

2. calculate z = ( p - 1 ) * ( q - 1 ) = ( 3 - 1 ) * ( 11 - 1 ) = 20

3. choose a prime number k, such that k is co-prime to z, i.e, z is not divisible by k. We have several choices for k: 7, 11, 13, 17, 19 (we cannot use 5, because 20 is divisible by 5).

Let's pick k=7 (smaller k, "less math").

4. the numbers n = 33 and k = 7 become the Server's public key.

5. the Server has to calculate it's secret key. Here is how.

6. k * j = 1 ( mod z )

7.)  7 * j = 1 ( mod 20 )

8.) ( 7 * j ) / 20 = ? with the remainder of 1 (the "?" here means: "something, but don't wory about it"; we are only interested in the remainder).

Since we selected (on purpose) to work with small numbers, we can easily conclude that 21 / 20 gives "something" with the remainder of 1. So, 7 * j = 21, and j = 3. This is our secret key. We MUST NOT give this key to anyone. Now, server sends n=33 and k=7 back to the Browser

Q 3. If a packet arrive with an M-bit value is ‘1’ and a fragmentation offset value ‘0’, then it is ______ fragment.
(A) First                                 (B) Middle                         (C) Last         (D) All of the above

Answer is A. If the M bit is 0, it means that there are no more fragments. If the M bit is 1, it means that there is at least one more fragment. This fragment can be the first one or a middle one, but not the last one. Offset value '0' always means this is the first fragment. To determine the position of the first byte in a particular fragment, we would multiply given offset value with 8. For example, offset is 200, then the position of first byte in that fragment is 1600.

Fragment Offset field solves the problem of sequencing fragments by indicating to the recipient device where in the overall message each particular fragment should be placed. The field is 13 bits wide, so the offset can be from 0 to 8191. Fragments are specified in units of 8 bytes, which is why fragment length must be a multiple of 8. Uncoincidentally, 8191 * 8 is 65,528, just about the maximum size allowed for an IP datagram.

IMP: When an MTU requirement forces a datagram to be fragmented, it is split into several smaller IP datagrams, each containing part of the original. The header of the original datagram is changed into the header of the first fragment, and new headers are created for the other fragments. Each is set to the same Identification value to mark them as part of the same original datagram. The Fragment Offset of each is set to the location where the fragment belongs in the original. The More Fragments field is set to 1 for all fragments but the last, to let the recipient know when it has received all the fragments.

Q 4. In a network of LANs connected by bridges, packets are sent from one LAN to another through intermediate bridges. Since more than one path may exist between two LANs, packets may have to be routed through multiple bridges. Why is the spanning tree algorithm used for bridge-routing? (GATE 2005)

(a) For shortest path routing between LANs         (b) For avoiding loops in the routing paths

(c) For fault tolerance                                      (d) For minimizing collisions

SOLUTION: Spanning tree is a protocol that allows the bridges to exchange information so that only one of them will handle a given message that is being sent between two computers within the network. This  protocol prevents the condition known as a BRIDGE LOOP. It is typical to add a second bridge between two network segments as a backup in case the primary bridge fails (both bridges need to have some way to understand which bridge is the primary one). To do this, they have a separate path connection just between the bridges in which they exchange information, using bridge protocol data units (BPDUs).

The program in each bridge that allows it to determine how to use the protocol is known as the spanning tree algorithm. The algorithm is specifically constructed to avoid bridge loops (for a bridge using only the most efficient path when faced with multiple paths). If the best path fails, the algorithm recalculates the network and finds the next best route.

Q 5. How many 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission with one start bit, eight data bits, two stop bits and one parity bit?
(1) 600        (2) 800        (3) 876        (4) 1200

For 9600 baud, 1 bit=1/9600=0.104mS. Each char would require 11 bits. That means, to transmit one char, it would take 1.144 mS. So, applying the formula 1000/1.144 (mS in one Sec/transmission time for one char)=approx 875. Nearest answer is (3)

Q 6. The single stage network is also called

A) one sided network                             B) two sided network

C) recirculating network                      D) pipeline network

Single-Stage Network is a single stage of switching elements (SEs) existing between the inputs and the outputs of the network. Data is circulated a number of times around the network.

Q 7. If a class B network on the Internet has a subnet mask of 255.255.248.0, what is the maximum number of hosts per subnet?
(a) 1022 (b) 1023 (c) 2046 (d) 2047

Explanation: Convert the subnet mask into binary format.

255.255.248.0 = 11111111.11111111.11111000.00000000

Number of 1's in the subnet mask indicates the Network-ID and the Subnet-ID part. Number of 0's in the subnet mask indicates the Host-ID part. Maximum number of Hosts per subnet = 2¹¹ = 2048, where 11 = Number of 0's in the Subnet Mask. Out of 2048 values, 2 addresses are reserved, hence we remove them (2048-2 = 2046). Note: In the host part of the address:- all bits as 1 is reserved as broadcast address and all bits as 0 is used as network address of subnet.

Q 8. Which of the following system calls results in the sending of SYN packets?(a) socket (b) bind (c) listen (d) connect

The connect system call is normally called by the client process to connect to the server process. The socket system call creates a new socket and assigns the protocol and resources to the created socket descriptor. The bind system call associates a local network transport address with a socket. For a client process, it is not mandatory to issue a bind call. The kernel takes care of doing an implicit binding when the client process issues the connect system call. 

It is often necessary for a server process to issue an explicit bind request before it can accept connections or start communication with clients. The listen call indicates to the protocol that the server process is ready to accept any new incoming connections on the socket. There is a limit on the number of connections that can be queued up, after which any further connection requests are ignored.

What is the maximum size of data that the application layer can pass on to the TCP layer below?
Q 9. (a) Any size (b) 2¹⁶ bytes-size of TCP header     (c) 2¹⁶ bytes   (d) 1500 bytes

Which of the following transport layer protocols is used to support electronic mail?

(a) SMTP (b) IP  (c) TCP  (d) UDP

Q 10. When a host on network A sends a message to a host on network B, which address does the router look at ?
(a) Port                  (b) IP           (c) Physical              (d) Subnet mask

Q 11. The ....................... field in the SNMP PDU, reports an error in a response message.

(A) Error index         (B) Error status       (C) Set request                  (D) Agent index

Error status: An integer value that is used in a Response-PDU to tell the requesting SNMP entity the result of its request. A value of zero indicates that no error occurred; the other values indicate what sort of error happened.

Error Index: When Error Status is non-zero, this field contains a pointer that specifies which object generated the error. Always zero in a request.

Notes on Data Communication & Networks :Part II

Facts on Data Communication & Networks:

Hello Friends!

Data Communication & Networks is one of the important subjects in UGC NET exam on which lots of questions are asked. So, prepare this subject well. The syllabus of the subject is vast, but is one of the easiest subjects to prepare. 

Please read the post and then post your comments/suggestions:

ETHERNET: Ethernet frames must carry a minimum payload of 46 bytes, which ensures that a valid Ethernet frame is 512-bits long (considering bits of header section also). The shortest Ethernet frame is 64 bytes in length, which carry Control messages.

Ethernet frames don't have a way to indicate end-of-frame, but an inter-frame gap (of time required to send 96 bit of data, i.e. 9.6 micro sec.) is used.

Horn antennas are very popular at UHF (300 MHz-3 GHz) and higher frequencies. They have a wide impedance bandwidth, implying that the input impedance is slowly varying over a wide frequency range.

S-parameters describe the input-output relationship between ports (or terminals) in an electrical system. A port can be defined as any place where we can deliver voltage and current.

Waveforms are not made up of a discrete number of frequencies, but rather a continuous range of frequencies (sum of frequencies).

Through modulation the waveforms can be relocated to separate frequency bands. Television is broadcast primarily at 54-216 MHz. FM radio operates between 87.5-108 MHz.

Frequency Band Name	Frequency Range	Wavelength (Meters)	Application
Extremely Low Frequency (ELF)	3-30 Hz	10,000-100,000 km	Underwater Communication
Super Low Frequency (SLF)	30-300 Hz	1,000-10,000 km	AC Power (though not a transmitted wave)
Ultra Low Frequency (ULF)	300-3000 Hz	100-1,000 km	Earth Mode Communications*
Very Low Frequency (VLF)	3-30 kHz	10-100 km	Navigational Beacons
Low Frequency (LF)	30-300 kHz	1-10 km	AM Radio
Medium Frequency (MF)	300-3000 kHz	100-1,000 m	Aviation and AM Radio
High Frequency (HF)	3-30 MHz	10-100 m	Shortwave Radio
Very High Frequency (VHF)	30-300 MHz	1-10 m	FM Radio
Ultra High Frequency (UHF)	300-3000 MHz	10-100 cm	Television, Mobile Phones, GPS
Super High Frequency (SHF)	3-30 GHz	1-10 cm	Satellite Links, Wireless Communication
Extremely High Frequency (EHF)	30-300 GHz	1-10 mm	Astronomy, Remote Sensing
Visible Spectrum	400-790 THz (4*10^14-7.9*10^14)	380-750 nm (nanometers)	Human Eye

* Communications through the ground using conduction fields e.g. military communications.

Slot Time: It is twice the time it takes for an electronic pulse to travel the maximum distance between two nodes. Thus Propagation delay takes half of the time of slot time since it is only the measure of the time required for signal to reach from node A to B. Slot time is used for half-duplex Ethernet network operation. It is 512 bit times for Ethernet networks operating below 1 Gbit/s, and 4096 bit times for Gigabit Ethernet. To reliably detect collisions, the minimum transmission time for a complete frame must be at least one slot time, whereas the round-trip propagation delay must be less than a slot time (half of slot time).

Back-off Algorithm: Once a collision is detected by simultaneous transmitters, they will follow a backoff algorithm, which requires each transmitter to wait an integral number of slot times (51.2 µs) before attempting a new transmission sequence. The integer is determined by the equation:
0<=r<2 power k where k = min (n, 10)
The variable k is actually the number of collisions capped at a maximum of 10. Therefore, r can range from 0 to 1023. The actual value for r is determined by a random process within each Ethernet node. As the number of consecutive collisions increases, the range of possible backoff times increases exponentially. The number of possible retries is max. 16.
There are no collisions with a full-duplex link, where each node is paired with a port on the hub.

Collision Domain - a section of a network where data packets can collide with one another when being sent on a shared medium or through repeaters, in particular.

The 5-4-3 rule: A system can have up to five segments in series, with up to four repeaters and no more than three mixing segments (a segment that may be connected to more than two transceivers).

The FCS field in Ethernet frame provides only bit-level error detection, no error recovery.

UDP is also known as laissez-faire protocol

TCP is used for unicast addresses only, so multicast applications must use the UDP transport protocol.

In asynchronous transmission, the Start bit always has a value of 0 (a Space). The Stop Bit always has a value of 1 (a Mark). This means that there will always be a Mark (1) to Space (0) transition on the line at the start of every word.

Application layer is free to send any size of data, there is no upper limit defined by standards. The lower layers divides the data if needed.
A channel with x bps may not necessarily transmit data at x rate, since protocols, encryption, and other factors can add may overheads.

The asymptotic bandwidth (formally asymptotic throughput) for a network is the measure of maximum throughput for a greedy source (a traffic generator that generates data at the maximum rate possible and at the earliest opportunity possible).

CIDR: Classless Inter-Domain Routing, known as supernetting, is a solution to limited address space problem in a network. It allocates address space to ISPs and end users on any address bit boundary, instead of on 8-bit segments (which is class based addressing). It appends to the IP address a slash character and the decimal number as routing prefix, e.g., "192.168.2.0/24" for IPv4, and 2001:db8::/32 for IPv6. The value after / indicates how many bits are used for the network prefix, leaving the remaining bits to identify the specific host.

CIDR currently uses prefixes anywhere from 13 to 27 bits. This solution fits an organization's specific needs. It helps in reducing number of entries in global routing tables. It is the concept of subnetting within the internet itself.

The industrial, scientific and medical (ISM) radio bands are radio bands (portions of the radio spectrum) reserved internationally for industrial, scientific and medical purposes other than communications. These are for unlicensed operations. Cordless phones, Bluetooth devices, near field communication (NFC) devices, and wireless computer networks all use frequencies allocated to low power communications as well as ISM.

Hartley's law- "The maximum data rate of a physical communication link is proportional to its bandwidth in hertz, which is sometimes called frequency bandwidth, spectral bandwidth, RF bandwidth, signal bandwidth or analog bandwidth."

A Baud Rate represents the number of bits that are actually being sent over the media, not the amount of data that is actually moved from one DTE device to the other. That means, baud rate decides the actual bit rate. For example, the bit rate is 9600

The Intelligent Network (IN) is the standard network architecture which allows telecom operators to differentiate themselves by providing value-added services in addition to the standard telecom services. The intelligence is provided by network nodes on the service layer (a conceptual layer within a network service provider architecture. It aims at providing middleware that serves third-party value-added services and applications at a higher application layer.)

The Internet protocol suite ( TCP/IP Model), occasionally known as the DoD model due to the foundational influence of the ARPANET. The TCP/IP model and related protocols are maintained by the Internet Engineering Task Force (IETF).

PORT: Each process that wants to communicate with another process identifies itself to the TCP/IP protocol suite by one or more ports. Application Layer talks with Transport layer through ports. A port number helps the transport layer protocols (like TCP) to know the type of content residing inside the packet.

A port is a 16-bit number, used by the host-to-host protocol to identify to which higher level protocol or application program (process) it must deliver incoming messages. There are two types of ports. Well-known port numbers(0-1023) are typically odd, because early systems using the port concept required an odd/even pair of ports for duplex operations.

The well-known ports are controlled and assigned by the Internet Assigned Number Authority (IANA) and on most systems can only be used by system processes or by programs executed by privileged users. Ephemeral ports are opposite to well-known ports. Such port number are used by clients and are contained in the UDP datagrams sent to the server.

Normally, a server will use either TCP or UDP, but there are exceptions. For example, domain name servers use both UDP port 53 and TCP port 53.

SOCKET: A socket is a special type of file handle, which is used by a process to request network services from the operating system. A socket address is the triple: <protocol, local-address, local-process>. For example, in the TCP/IP suite:

<tcp, 193.44.234.3, 12345>

An association is the 5-tuple that completely specifies the two processes that comprise a connection:

<protocol, local-address, local-process, foreign-address, foreign-process>. In the TCP/IP suite, the following could be a valid association:

<tcp, 193.44.234.3, 1500, 193.44.234.5, 21>

Two processes communicate via TCP sockets. The socket model provides a process with a full-duplex byte stream connection to another process.

UDP: UDP is basically an application interface to IP. It adds no reliability, flow-control, or error recovery to IP. It simply serves as a multiplexer/demultiplexer for sending and receiving datagrams, using ports to direct the datagrams. It requires the application to take responsibility for error recovery and so on.

Slow-start is one of the algorithms that TCP uses to control congestion inside the network. It is also known as the exponential growth phase.

Broadband means "having instantaneous bandwidths greater than 1 MHz and supporting data rates greater than about 1.5 Mbit/s." In telecommunication, Broadband refers to a communication bandwidth of at least 256 kbit/s. Each channel is 6 MHz wide.

The additive-increase/multiplicative-decrease (AIMD) algorithm is a feedback control algorithm used by  TCP for Congestion Avoidance. The approach taken is to increase the transmission rate (window size), probing for usable bandwidth, until loss occurs.

IP address classes:

Class	Leftmost bits	Start address	Finish address
A	0xxx	0.0.0.0	127.255.255.255
B	10xx	128.0.0.0	191.255.255.255
C	110x	192.0.0.0	223.255.255.255
D	1110	224.0.0.0	239.255.255.255
E	1111	240.0.0.0	255.255.255.255

IP address range for Intranets (Private Networks):

Class	Private start address	Private finish address
A	10.0.0.0	10.255.255.255
B	172.16.0.0	172.31.255.255
C	192.168.0.0	192.168.255.255

IP packets addressed by them cannot be transmitted onto the public Internet. If such a private network needs to connect to the Internet, it must use either a network address translator (NAT) gateway, or a proxy server.In IPV6, The address block fc00::/7 has been reserved for private networks.

IP officially reserves the entire range from 127.0.0.0 through 127.255.255.255 for loopback purposes.

IPv6 does not use classes. IPv6 supports the following three IP address types: unicast, multicast and anycast. IPv6 does not support broadcast. Multicast addresses in IPv6 start with 'FF' (255) just like IPv4 addresses. Unicast addresses have 3 defined scopes, including link-local, site-local and global; and multicast addresses have 14 scopes.

The number of IPv6 addresses is 10^28. There is no ARP in V6. Currently, DHCP, FTP, PPP, RIP, SNMP, VPN, L2TP and Telnet do not support IPv6.

IPv6 does not require NAT. NAT, too, doesn't support V6. Currently, IPv6 packets are not forwarded.

IPv6 reserves just two special addresses: 0:0:0:0:0:0:0:0 and 0:0:0:0:0:0:0:1. IPv6 uses 0:0:0:0:0:0:0:0 internal to the protocol implementation, so nodes cannot use it for their own communication purposes. IPv6 uses 0:0:0:0:0:0:0:1 as its loopback address, equivalent to 127.0.0.1 in IPv4. The minimum size of an IP datagram is 28 bytes, including 20 bytes of header.

BIND and NSD are the most widely used DNS software on the Internet

Anycast is a network addressing and routing methodology in which datagrams from a single sender are routed to the topologically nearest node in a group of potential receivers, though it may be sent to several nodes, all identified by the same destination address. On the Internet, anycast is usually implemented by using BGP.

In denial-of-service attacks, a rogue network host may advertise itself as an anycast server for a vital network service, to provide false information or simply block service.

"6 to 4" is an Internet transition mechanism for migrating from IPv4 to IPv6, a system that allows IPv6 packets to be transmitted over an IPv4 network. 6to4 does not facilitate interoperation between IPv4-only hosts and IPv6-only hosts, but simply a transparent mechanism used as a transport layer between IPv6 nodes.

The network requests supporting DNS lookups run over TCP and UDP, port 53 by default.